Arthmetic

Aim:
Write 100 as the sum of two integers, one divisible by 7 and the other divisible by 11. Use
your answer to find formulas giving all the solutions of the following equation where x and
y are integers.

7x+11y=100

Working out:

Here I listed the multiples of 7 plus the number they need to make a 100.
I highlighted 56+44=100 because from all, 44 is the only number which is multiple of 11

7+93=100
14+86=100
21+79=100
28+72=100
35+65=100
42+58=100
49+51=100
56+44=100
63+37=100
70+30=100
77+23=100
84+16=100
91+9=100
98+2=100

7x8=56 and 11x4=44 so 7(8)+11(4)

Now I'm gonna try it, this time with negative numbers
-7+107=100
-14+114=100
-21+121=100
-28+128=100
-35+135=100
-42+142=100
-49+149=100
-56+156=100
-63+163=100
-70+170=100
-77+177=100
-84+184=100
-91+191=100
-98+198=100
-105+205=100
-112+212=100
-119+219=100
-126+226=100
-133+233=100
-147+247=100
-154+254=100
-161+261=100
-168+268=100
-175+275=100


X Y
30 -10
19 -3
8 4
-3 11
-14 18
-25 25

the difference between the numbers in X is 11 (eg. 30-19=11, 19-8=11...)
And the difference between the numbers in Y is 7(eg. -10- -3= 7, -3 - 4=7...)
Now Im Using 8 and 4 because, from all the numbers, it's the only time that both numbers are positive. I'ts easier

Rule:
X= -11n+19
Y= 7n-3

Checking:

1. X: -11(3)+19=-14
Y: 7(3)-3=18
7(-14)=-98
11(18)=198
-98+198=100

2. X:-11(7)+19=-58
Y: 7(7)-3=46
7(-58)=-406
11(46)=506
-406+506=100

as you can see, IT WORKS! :)